Increasing/Decreasing & Local Extrema - Business Calculus

The profit $P(x)$ in dollars from manufacturing and selling $x$ units of a product is given by $P(x) = -0.15x^2 + 480x - 60000$ . The company's production facility can operate at any level $x \ge 0$ . For which interval of production is the company's profit increasing?
Answer: For $x$ in $(0, 1600)$ To find where the profit function is increasing, we need to find the interval where its derivative is positive. The profit function is $P(x) = -0.15x^2 + 480x - 60000$ . The derivative is $P'(x) = -0.3x + 480$ . We set $P'(x) > 0$ to find where profit is increasing: $-0.3x + 480 > 0 \implies 480 > 0.3x \implies 1600 > x$ . Since production must be non-negative, the interval is $(0, 1600)$ .
The total revenue $R(x)$ in thousands of dollars from selling $x$ hundred units of a product is given by $R(x) = \frac{1}{3}x^3 - 5x^2 + 25x + 10$ for $x \ge 0$ . A marketing analyst claims that there is a production level $x>0$ where revenue reaches a local maximum. Which statement correctly evaluates this claim?
Answer: The claim is false; the revenue function is always increasing for $x \ge 0$ and has no local extrema. To find local extrema, we analyze the first derivative: $R'(x) = x^2 - 10x + 25 = (x-5)^2$ . The only critical point occurs when $R'(x)=0$ , which is at $x=5$ . However, since $R'(x)=(x-5)^2$ is a square, $R'(x) \ge 0$ for all $x$ . The derivative is positive for $x \neq 5$ and zero at $x=5$ . Because the derivative never becomes negative, the function never decreases. Therefore, there is no local maximum or minimum. The function is always increasing (or momentarily stationary), so the analyst's claim is false.
A company's revenue from selling $x$ units of a product is given by $R(x) = \frac{5000x}{x+20}$ . Which of the following statements is true regarding the company's revenue?
Answer: Revenue is always increasing, but the rate of increase slows as production grows. To determine where revenue is increasing or decreasing, we examine the first derivative, $R'(x)$ . Using the quotient rule: $R'(x) = \frac{5000(x+20) - 5000x(1)}{(x+20)^2} = \frac{5000x + 100000 - 5000x}{(x+20)^2} = \frac{100000}{(x+20)^2}$ . For any production level $x \ge 0$ , the numerator is positive and the denominator is positive. Thus, $R'(x)$ is always positive. This means the revenue function $R(x)$ is always increasing for $x \ge 0$ . To analyze the rate of increase, we look at the second derivative, $R''(x)$ . $R''(x) = -2 \cdot 100000 (x+20)^{-3} = \frac{-200000}{(x+20)^3}$ . For $x \ge 0$ , $R''(x)$ is always negative, which means the function is concave down. This indicates that while revenue is always increasing, the rate of that increase is slowing down.
A subscription service's monthly revenue function is $R(m) = m^4 - 6m^3 + 9m^2 + 100$ thousands of dollars, where $m$ is months since launch. The CFO needs to determine the intervals where revenue is decreasing to plan cash flow. On which interval(s) is revenue decreasing?
Answer: $m \in (1, 3)$ only, since revenue decreases between the critical points To find where revenue is decreasing, solve $R'(m) < 0$ . First, $R'(m) = 4m^3 - 18m^2 + 18m = 2m(2m^2 - 9m + 9)$ . For the quadratic factor, use the quadratic formula: $m = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm 3}{4}$ . So $m = 3$ or $m = \frac{3}{2}$ . Therefore, $R'(m) = 2m(2m - 3)(m - 3)$ . Actually, let me recalculate: $2m^2 - 9m + 9 = 0$ gives $m = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm 3}{4}$ , so $m = 3$ or $m = \frac{3}{2}$ . But this means $2m^2 - 9m + 9 = 2(m - 3)(m - \frac{3}{2}) = (m - 3)(2m - 3)$ . So $R'(m) = 2m(m - 3)(2m - 3)$ . The critical points are $m = 0, \frac{3}{2}, 3$ . To determine the sign of $R'(m)$ in each interval: - For $0 < m < \frac{3}{2}$ : $R'(m) = (+)(-)(-) = (+) > 0$ (increasing) - For $\frac{3}{2} < m < 3$ : $R'(m) = (+)(-)(+) = (-) < 0$ (decreasing) - For $m > 3$ : $R'(m) = (+)(+)(+) = (+) > 0$ (increasing) Therefore, revenue is decreasing on $(\frac{3}{2}, 3)$ . Since $\frac{3}{2} = 1.5$ , this is approximately the interval $(1, 3)$ as given in option B.
A company can produce a maximum of 100 widgets per day. The daily profit, $P(x)$ , from producing $x$ widgets is given by $P(x) = -x^3 + 150x^2 - 4800x$ for the domain $x \in [0, 100]$ . On which interval(s) of production is the profit decreasing?
Answer: On the intervals $[0, 20)$ and $(80, 100]$ Profit is decreasing where its derivative, $P'(x)$ , is negative. First, find the derivative: $P'(x) = -3x^2 + 300x - 4800$ . To find where $P'(x) < 0$ , we first find the roots of $P'(x)=0$ : $-3(x^2 - 100x + 1600) = 0 \implies -3(x-20)(x-80)=0$ . The critical points are $x=20$ and $x=80$ . Since $P'(x)$ is a downward-opening parabola, it is negative outside its roots, i.e., for $x < 20$ or $x > 80$ . We must consider the function's domain, $[0, 100]$ . Intersecting $(-\infty, 20) \cup (80, \infty)$ with $[0, 100]$ gives the intervals $[0, 20)$ and $(80, 100]$ .
A company's marginal revenue is given by $R'(x) = 150 - 4x$ and its marginal cost is $C'(x) = 2x + 30$ , where $x$ is the number of units produced. For what range of production levels will an increase in production result in an increase in profit?
Answer: For $x < 20$ Profit $P(x)$ increases when the marginal profit $P'(x)$ is positive. The marginal profit is the difference between marginal revenue and marginal cost: $P'(x) = R'(x) - C'(x)$ . Substituting the given functions: $P'(x) = (150 - 4x) - (2x + 30) = 150 - 4x - 2x - 30 = 120 - 6x$ . To find where profit is increasing, we set $P'(x) > 0$ : $120 - 6x > 0$ . Solving for $x$ : $120 > 6x$ , which simplifies to $20 > x$ . Therefore, profit increases for production levels less than 20 units.
The productivity $V(t)$ of a worker $t$ hours into an 8-hour shift, for $t \in [0, 8]$ , is modeled by $V(t) = t^4 - 16t^3 + 72t^2 + 10$ . At what time(s) $t$ during the shift do local minima of productivity occur?
Answer: At $t=0$ only When you encounter a question about local minima of a function, you need to find critical points by taking the derivative and then use the second derivative test to classify them. To find where local minima occur, start by finding the critical points. Take the derivative: $V'(t) = 4t^3 - 48t^2 + 144t = 4t(t^2 - 12t + 36) = 4t(t-6)^2$ . Setting this equal to zero gives critical points at $t = 0$ and $t = 6$ . Next, use the second derivative test to classify these points. The second derivative is $V''(t) = 12t^2 - 96t + 144$ . At $t = 0$ : $V''(0) = 144 > 0$ , indicating a local minimum. At $t = 6$ : $V''(6) = 12(36) - 96(6) + 144 = 432 - 576 + 144 = 0$ . Since the second derivative is zero, this test is inconclusive, but examining the first derivative shows $V'(t) = 4t(t-6)^2 \geq 0$ for all $t \geq 0$ , meaning the function is non-decreasing. At $t = 6$ , there's a horizontal inflection point, not a local minimum. Choice A incorrectly includes $t = 6$ , which is an inflection point, not a minimum. Choice B only includes $t = 6$ , missing the actual minimum at $t = 0$ and incorrectly classifying the inflection point. Choice D incorrectly includes $t = 8$ , which is an endpoint, not a critical point. Therefore, C is correct: the local minimum occurs only at $t = 0$ . Remember: always verify critical points with the second derivative test, and distinguish between local extrema and inflection points where the second derivative equals zero.
The daily profit for a company, in hundreds of dollars, is modeled by $P(x) = \frac{1}{4}x^4 - 2x^3 + 4x^2 - 5$ , where $x$ represents thousands of units produced. The company's operations are stable only when production is at a level corresponding to a local maximum of profit. What is this production level?
Answer: $x = 2$ To find local extrema, we first compute the derivative of the profit function: $P'(x) = x^3 - 6x^2 + 8x$ . We find the critical points by setting $P'(x)=0$ : $x(x^2 - 6x + 8) = 0$ , which factors as $x(x-2)(x-4) = 0$ . The critical points are $x=0$ , $x=2$ , and $x=4$ . We use the first derivative test to classify them. We check the sign of $P'(x)$ in the intervals defined by these points: $(-\infty, 0)$ , $(0, 2)$ , $(2, 4)$ , and $(4, \infty)$ . Let's use test points: $P'(-1)=-15$ (decreasing), $P'(1)=3$ (increasing), $P'(3)=-3$ (decreasing), $P'(5)=15$ (increasing). At $x=0$ , $P'$ changes from - to +, so it's a local minimum. At $x=2$ , $P'$ changes from + to -, so it's a local maximum. At $x=4$ , $P'$ changes from - to +, so it's a local minimum. The only local maximum occurs at $x=2$ , which corresponds to 2,000 units.
After a successful advertising campaign, the monthly profit $P$ (in thousands of dollars) for a product is modeled by $P(t) = \frac{120t}{t^2 + 36}$ , where $t$ is the number of months after the campaign ended ( $t \ge 0$ ). The positive effects of the campaign eventually wear off and profit begins to decline. After how many months does this decline begin?
Answer: 6 months The profit begins to decline when the profit function $P(t)$ reaches its maximum and starts decreasing. This occurs at a critical point. We find the derivative of $P(t)$ using the quotient rule: $P'(t) = \frac{120(t^2 + 36) - 120t(2t)}{(t^2 + 36)^2}$ . To find critical points, we set the numerator to zero: $120(t^2 + 36) - 240t^2 = 0$ . This simplifies to $120t^2 + 120 \cdot 36 - 240t^2 = 0$ , which gives $120 \cdot 36 - 120t^2 = 0$ . Dividing by 120, we get $36 - t^2 = 0$ , so $t^2 = 36$ . Since $t \ge 0$ , we have $t=6$ . For $0 \le t < 6$ , $P'(t)$ is positive (profit increasing), and for $t > 6$ , $P'(t)$ is negative (profit decreasing). Therefore, the decline in profit begins after 6 months.
The daily cost to a factory for producing $x$ items is $C(x) = 0.001x^2 - kx + 15000$ , where $k$ is a positive constant representing an efficiency factor. It is known that the cost function has a local minimum at a production level of $x=500$ items. What is the value of the efficiency factor $k$ ?
Answer: $k = 1$ For a local minimum to occur at $x=500$ , the first derivative of the cost function must be zero at that point. The derivative is $C'(x) = 0.002x - k$ . Setting $C'(500) = 0$ , we get $0.002(500) - k = 0$ , which simplifies to $1 - k = 0$ . Therefore, $k=1$ . To confirm it's a minimum, the second derivative is $C''(x) = 0.002$ , which is positive, confirming that any critical point is a local minimum.
A manufacturing company's cost function is $C(x) = x^4 - 8x^3 + 18x^2 - 8x + 100$ dollars for producing $x$ thousand units. The production manager needs to identify all local minimum points to optimize costs. How many local minima does this cost function have?
Answer: Two local minima, since the quartic function has multiple turning points To find local minima, first find critical points by setting $C'(x) = 0$ . $C'(x) = 4x^3 - 24x^2 + 36x - 8$ . This factors as $C'(x) = 4(x^3 - 6x^2 + 9x - 2)$ . By trying rational roots, we find that $x = 2$ is a root: $8 - 24 + 18 - 2 = 0$ . Factoring: $x^3 - 6x^2 + 9x - 2 = (x-2)(x^2 - 4x + 1)$ . Using the quadratic formula on $x^2 - 4x + 1 = 0$ : $x = 2 \pm \sqrt{3}$ . So the critical points are $x = 2 - \sqrt{3} \approx 0.27$ , $x = 2$ , and $x = 2 + \sqrt{3} \approx 3.73$ . To determine which are local minima, use the second derivative test: $C''(x) = 12x^2 - 48x + 36 = 12(x^2 - 4x + 3) = 12(x-1)(x-3)$ . At $x = 2 - \sqrt{3} \approx 0.27$ : $C''(0.27) = 12(0.27-1)(0.27-3) = 12(-0.73)(-2.73) > 0$ , so this is a local minimum. At $x = 2$ : $C''(2) = 12(1)(-1) = -12 < 0$ , so this is a local maximum. At $x = 2 + \sqrt{3} \approx 3.73$ : $C''(3.73) = 12(2.73)(0.73) > 0$ , so this is a local minimum. Therefore, there are two local minima.
The total cost in dollars to produce $x$ units of a certain item is $C(x) = x^2 + 10x + 1600$ . For $x > 0$ , on which production interval is the average cost per unit decreasing?
Answer: On the interval $(0, 40)$ The average cost function is $AC(x) = C(x)/x = x + 10 + 1600/x$ . To find where the average cost is decreasing, we must find where its derivative is negative. The derivative is $AC'(x) = 1 - 1600/x^2$ . Set $AC'(x) < 0$ : $1 - 1600/x^2 < 0 \implies 1 < 1600/x^2 \implies x^2 < 1600$ . Since $x > 0$ , this gives $0 < x < 40$ . Thus, the average cost is decreasing on the interval $(0, 40)$ .
A renewable energy company's power output function is $P(w) = w^3 - 15w^2 + 63w + 50$ kilowatts, where $w$ represents wind speed in m/s. Engineers need to determine the wind speed intervals where power output is decreasing to optimize turbine settings. For which wind speeds is power output decreasing?
Answer: $w \in (3, 7)$ m/s, where the derivative is negative between critical points When you encounter a question about finding where a function is decreasing, you need to analyze the derivative and determine where it's negative. For power output functions like this one, decreasing intervals represent wind speeds where additional wind actually reduces power generation. To find where $P(w) = w^3 - 15w^2 + 63w + 50$ is decreasing, first calculate the derivative: $P'(w) = 3w^2 - 30w + 63$ . The function decreases when $P'(w) < 0$ . Setting $P'(w) = 0$ to find critical points: $3w^2 - 30w + 63 = 0$ , which simplifies to $w^2 - 10w + 21 = 0$ . Factoring gives $(w-3)(w-7) = 0$ , so the critical points are $w = 3$ and $w = 7$ . Since $P'(w)$ is a upward-opening parabola (positive leading coefficient), it's negative between the roots. Testing $w = 5$ : $P'(5) = 75 - 150 + 63 = -12 < 0$ . Therefore, $P'(w) < 0$ for $w \in (3,7)$ . Answer A gives the wrong interval $(5,9)$ - while 5 is in the decreasing region, the function starts increasing again after $w = 7$ . Answer B incorrectly includes $(0,3)$ and $(7,∞)$ where the derivative is actually positive. Answer D is wrong because the derivative calculation clearly shows decreasing intervals exist. Study tip: Always find critical points by setting the derivative equal to zero, then test intervals between critical points to determine where the function increases or decreases. The sign of the derivative tells the whole story.
The value of a rare collectible, in dollars, is modeled by $V(t) = 50t e^{-0.1t}$ where $t$ is the number of years from today. To maximize the value, for how many years should the owner hold onto the collectible before selling it?
Answer: $10$ years To find the maximum value, we need to find the critical points of the value function $V(t)$ by taking its derivative and setting it to zero. Using the product rule, $(uv)' = u'v + uv'$ , let $u = 50t$ and $v = e^{-0.1t}$ . Then $u' = 50$ and $v' = -0.1e^{-0.1t}$ . So, $V'(t) = 50e^{-0.1t} + 50t(-0.1e^{-0.1t}) = 50e^{-0.1t}(1 - 0.1t)$ . Set $V'(t) = 0$ . Since $50e^{-0.1t}$ is always positive, we only need to solve $1 - 0.1t = 0$ . This gives $0.1t = 1$ , so $t = 10$ . The first derivative test shows that for $t < 10$ , $V'(t) > 0$ (increasing) and for $t > 10$ , $V'(t) < 0$ (decreasing), confirming a local maximum at $t=10$ years.
An online platform's user engagement function is $E(h) = h^3 - 9h^2 + 24h + 10$ where $h$ represents hours after a content update. The platform's algorithm needs to identify when engagement is at a local maximum to schedule the next update optimally. At which hour does the first local maximum occur after the update?
Answer: $h = 2$ hours, where engagement peaks before declining significantly To find local maxima, first find critical points by setting $E'(h) = 0$ . $E'(h) = 3h^2 - 18h + 24 = 3(h^2 - 6h + 8) = 3(h-2)(h-4)$ . So the critical points are $h = 2$ and $h = 4$ . Use the second derivative test: $E''(h) = 6h - 18 = 6(h - 3)$ . At $h = 2$ : $E''(2) = 6(2-3) = -6 < 0$ , so $h = 2$ is a local maximum. At $h = 4$ : $E''(4) = 6(4-3) = 6 > 0$ , so $h = 4$ is a local minimum. Since we want the first local maximum after the update ( $h = 0$ ), the answer is $h = 2$ .
The profit, in thousands of dollars, from producing $x$ hundred units of a product is given by the function $P(x) = -x^3 + 15x^2 - 48x + 30$ . The marketing department recommends that production should be in a range where profit is increasing. Which of the following production ranges meets this criterion?
Answer: Between 200 and 800 units To find where the profit function $P(x)$ is increasing, we need to find where its derivative, $P'(x)$ , is positive. First, find the derivative: $P'(x) = -3x^2 + 30x - 48$ . To find the critical points, set $P'(x) = 0$ : $-3(x^2 - 10x + 16) = 0$ , which simplifies to $-3(x-2)(x-8) = 0$ . The critical points are $x=2$ and $x=8$ . Since $P'(x)$ is a downward-opening parabola, it will be positive between its roots. Testing the intervals: for $x<2$ , $P'(x)<0$ ; for $2<x<8$ , $P'(x)>0$ ; for $x>8$ , $P'(x)<0$ . Therefore, profit is increasing on the interval $(2, 8)$ . Since $x$ is in hundreds of units, this corresponds to the range between 200 and 800 units.
The demand function for a product is given by $p = \sqrt{450 - q}$ , where $p$ is the price per unit and $q$ is the number of units demanded, for $0 \le q \le 450$ . The revenue is given by $R(q) = p \cdot q$ . For which interval of quantity $q$ is the revenue function increasing?
Answer: For $q$ in $[0, 300)$ When you see a question asking where a function is increasing, you need to find where its derivative is positive. This tests your understanding of the relationship between derivatives and function behavior. First, let's find the revenue function. Since $R(q) = p \cdot q$ and $p = \sqrt{450 - q}$ , we have $R(q) = q\sqrt{450 - q}$ . To determine where revenue is increasing, we need $R'(q) > 0$ . Using the product rule: $R'(q) = \sqrt{450 - q} + q \cdot \frac{-1}{2\sqrt{450 - q}} = \sqrt{450 - q} - \frac{q}{2\sqrt{450 - q}}$ Finding a common denominator: $R'(q) = \frac{2(450 - q) - q}{2\sqrt{450 - q}} = \frac{900 - 3q}{2\sqrt{450 - q}}$ For $R'(q) > 0$ , we need the numerator positive (the denominator is always positive for valid $q$ values): $900 - 3q > 0$ $900 > 3q$ $q < 300$ Combined with the domain constraint $q \geq 0$ , revenue increases for $q \in [0, 300)$ . Choice A is wrong because revenue actually decreases for $q > 300$ . Choice B incorrectly identifies the decreasing interval as the increasing one. Choice C uses $q = 225$ as the boundary, which might come from incorrectly setting $450 - q = q$ instead of properly finding where $R'(q) = 0$ . The correct answer is D: $[0, 300)$ . Study tip: Always find the derivative first when analyzing where functions increase or decrease. Set the derivative equal to zero to find critical points, then test intervals to determine the sign.
The profit, $P(t)$ , in millions of dollars, from a new technology is modeled by $P(t) = 20 - (t-9)^{2/3}$ for its first 17 years of use, i.e., on the interval $0 \le t \le 17$ . Which statement accurately describes a key feature of this profit model?
Answer: Profit reaches a local maximum at $t=9$ years, where its rate of change is undefined. To analyze the function $P(t) = 20 - (t-9)^{2/3}$ , we find its derivative: $P'(t) = -\frac{2}{3}(t-9)^{-1/3} = \frac{-2}{3\sqrt[3]{t-9}}$ . The derivative is never zero. However, $P'(t)$ is undefined at $t=9$ , which makes $t=9$ a critical point. We test the sign of $P'(t)$ around this point. For $t<9$ , $t-9$ is negative, so $\sqrt[3]{t-9}$ is negative, making $P'(t)$ positive. Thus, profit is increasing on $(0, 9)$ . For $t>9$ , $t-9$ is positive, making $P'(t)$ negative. Thus, profit is decreasing on $(9, 17)$ . Since the function changes from increasing to decreasing at $t=9$ , it has a local maximum at that point.
A logistics company's delivery efficiency function is $D(t) = \frac{t^3 - 6t^2 + 9t + 20}{t + 1}$ where $t$ represents weeks since implementing new routing software. The operations manager needs to determine when efficiency is increasing to evaluate the software's performance. For what values of $t > 0$ is delivery efficiency increasing?
Answer: $t \in (0, 1) \cup (5, \infty)$ where the numerator's derivative dominates the analysis When you encounter a question about when a function is increasing, you need to find where its derivative is positive. This requires applying the quotient rule to differentiate the rational function. To find $D'(t)$ , use the quotient rule: $\frac{d}{dt}\left[\frac{f(t)}{g(t)}\right] = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}$ Here, $f(t) = t^3 - 6t^2 + 9t + 20$ and $g(t) = t + 1$ , so $f'(t) = 3t^2 - 12t + 9$ and $g'(t) = 1$ . $D'(t) = \frac{(3t^2 - 12t + 9)(t + 1) - (t^3 - 6t^2 + 9t + 20)(1)}{(t + 1)^2}$ Expanding the numerator: $(3t^2 - 12t + 9)(t + 1) = 3t^3 - 9t^2 - 3t + 9$ So: $D'(t) = \frac{3t^3 - 9t^2 - 3t + 9 - t^3 + 6t^2 - 9t - 20}{(t + 1)^2} = \frac{2t^3 - 3t^2 - 12t - 11}{(t + 1)^2}$ Since $(t + 1)^2 > 0$ for all $t > 0$ , you need $2t^3 - 3t^2 - 12t - 11 > 0$ . Testing values shows this cubic is positive when $t \in (0, 1) \cup (5, \infty)$ . Answer D is correct. Answer A misses the interval $(0, 1)$ . Answer B incorrectly identifies the intervals where $D'(t) > 0$ . Answer C gives an interval where $D'(t) < 0$ . Remember: for rational functions, always apply the quotient rule carefully and factor the resulting numerator to find where the derivative changes sign.
Let $C(x)$ be the total cost function and $AC(x)$ be the average cost function for producing $x$ units of a good. If it is known that for production levels in the interval $(a, b)$ , the marginal cost is greater than the average cost (i.e., $C'(x) > AC(x)$ ), what can be concluded about the average cost function on this interval?
Answer: The average cost function is increasing on $(a,b)$ . This question tests your understanding of the relationship between marginal cost and average cost, which is fundamental in business calculus for analyzing production efficiency. To solve this, you need to find what happens to the average cost function when marginal cost exceeds average cost. Since $AC(x) = \frac{C(x)}{x}$ , you can find the derivative: $AC'(x) = \frac{xC'(x) - C(x)}{x^2} = \frac{C'(x) - AC(x)}{x}$ Given that $C'(x) > AC(x)$ on interval $(a,b)$ , this means $C'(x) - AC(x) > 0$ . Since $x > 0$ (you can't produce negative units), we have $AC'(x) = \frac{C'(x) - AC(x)}{x} > 0$ . When a function's derivative is positive, the function is increasing. Choice A is wrong because marginal cost being greater than average cost tells us nothing about whether total cost is decreasing. In fact, total cost typically increases with production. Choice B is incorrect because we just showed that $AC'(x) > 0$ , meaning average cost is increasing, not decreasing. Choice C is wrong because if average cost has a local minimum on $(a,b)$ , its derivative would need to equal zero at some point, but we established $AC'(x) > 0$ throughout the interval. Choice D is correct because $AC'(x) > 0$ means the average cost function is increasing on $(a,b)$ . Study tip: Remember that when marginal cost exceeds average cost, it "pulls up" the average, making average cost increase. This relationship appears frequently on business calculus exams.

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